In calculus, a line integral is represented as an integral in which a function is to be integrated along a curve. A line integral is also known as a path integral, curvilinear integral or a curve integral. Line integrals have several applications such as in electromagnetic, line integral is used to estimate the work done on a charged particle traveling along some curve in a force field defined by a vector field. In classical mechanics, line integral is used to compute the word performed on mass m moving in a gravitational field. In this article, we will study a line integral, line integral of a vector field, line integral formulas etc.

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A line integral is integral in which function to be integrated along some curve in the coordinate system. The function which is to be integrated can either be represented as a scalar field or vector field. We can integrate both scalar-valued function and vector-valued function along a curve. The value of the vector line integral can be evaluated by summing up all the values of the points on the vector field.

A line integral (also known as path integral) is an integral of some function along with a curve. One can also incorporate a scalar-value function along a curve, obtaining such as the mass of wire from its density. We can also incorporate certain types of vector-valued functions along a curve. These vector- valued functions are the ones whose input and output size are similar and we usually define them as vector fields.

The line integral of the vector field is also interpreted as the amount of work that a force field does on a particle as it moves along a curve.

The line integral example given below helps you to understand the concept clearly.

1. Find the line integral

\[\int_{c} (1 + x^{2}y)ds\]

Where C is considered as an ellipse

r(t) = 2( cos t) + (3 sin t)

for 0 ≤ t ≤ 2π

Solution:

We calculate,

ds = \[\sqrt{(-2 sint)^{2} + (3 cost)^{2}} dt\] = \[\sqrt{4 sin^{2}t + 9 cos^{2}t}\]

We have the integral

\[\int_{a}^{2\pi}\](1+ (2 cos t)²( 3 sin t) \[\sqrt{4 sin^{2}t + 9 cos^{2}t}\] dt.

Hence, we get line integral = 15.87

2. Evaluate \[\int_{c}\]4x³ ds where C is the line segment from (1,2) to (-2,-1).

Solution: Here is the parameterization of the curve

\[\overline{r}\](t) = (1-t) (1, 2) + t (-2,-1)

( 1-3t, 2 - 3t)

For, 0≤ t ≤ 1.

Note: we are changing the direction of the curve and this will also change the parameterization of the curve so we can ensure that we start/end at the proper point.

Here , you can see the line integral

\[\int_{c}\]4x³ ds = \[\int_{0}^{1}\]4(1-3t)³ \[\sqrt{9 + 9}\]dt

= 12\[\sqrt{2}\](-1/12) (1-3t)⁴\[\int_{0}^{1}\]

= 12\[\sqrt{2}\](-5/4)

-15\[\sqrt{2}\] = -21.213

1. The line integral is used to calculate

Force

Length

Area

Volume

2. The integral form of potential and field relation is given by the line integral.

True

False

FAQ (Frequently Asked Questions)

1. What are the Applications of the Line Integral?

A line integral has multiple applications. It is used to calculate the surface area of three-dimensional shapes. Sole of the line integral application in vector calculus is:

A line integral is used to calculate the magnitude of wire.

A line integral is used to calculate the inertia moment and center of the magnitude of wire.

It is used to calculate the magnetic field around a conductor in Ampere's law.

A line integral enables us to examine the voltage generated in a loop in Faraday's law of magnetic induction.

It is used to compute the work performed by a force on moving objects in a vector field.

2. What are the Line Vectors of the Scalar Field and the Vector Field Formula?

The line integrals formulas for the scalar field and vector field are given below:

Line integral formula for the scalar field

For a line integral of the scalar field with function f: U ⊆ → Kₙ, a line integral along with some smooth curve, C ⊂ U is represented as,

∫_{C} f(k) dx =∫_{a}^{b} f[k(t)] . k’(t)| dt

Here k’: [x, y] → z is an arbitrary parameterization of the curve.

k (a) and k(b) obtains the endpoints of Z and x < y.

Line integral formula for vector field

For a line integral of vector field with function f: U ⊆ → K_{n}, a line integral along with some smooth curve in the direction ’k’ C ⊂ U is represented as,

∫_{C} F(k) dx =∫_{a} F[k(t)].|k’(t)| dt

Here, “.” denotes the dot product.